Question 332: Electromagnetic Induction

Question

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure.
The potential difference developed across the ring when its speed is $v$, is
![](https://neetcare-image.cdn.examgoal.net/1qpqdn5k7tcgd56/a1071458-054d-40ec-b43e-1516943ac4f6/c773aea0-66e6-11ea-997e-790b4402a570/file-1qpqdn5k7tcgd57-320w.jpg)

Accepted Answer

Potential difference that is developed across ring when its speed is v :
$\varepsilon $ = B v (l
i
Ã¢ÂÂ l
f
)
where, l
i
Ã¢ÂÂ l
f
= displacement between end of semicircular ring = 2r
Hence, $\varepsilon $ = Bv(2r) where R is at high potential.

Answer

zero

Answer

${{Bv\pi {r^2}} \over 2}$ and P is at higher potential

Answer

$\pi rBV$ and R is at higher potential

Answer

2rBV and R is at higher potential

Electromagnetic Induction

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is v, is

Choose the correct answer: